Forum: 396 – Topic: Quick question about the notation in the homework

Correct.

$\Phi_{1,3}$ is the field and $|\Phi_{1,3}\rangle$ the corresponding state, which is a Virasoro primary with a null vector at level $1\times3=3$.

The problem is to find an explicit expression for that null vector (as a linear combination of $L_{-3}|\Phi_{1,3}\rangle$, $L_{-2}L_{-1}|\Phi_{1,3}\rangle$ and $L_{-1}^3|\Phi_{1,3}\rangle$).

Forum: 375R – Topic: Welcome

Here’s a place to discuss Physics 375R. You need to sign up (by clicking on the link at the upper left), before you can post.

Equations are pretty easy to type. If you type

The position, $x(t)$, is
$$
  x(t) = x_0 + v_0 t + \frac{1}{2} a t^2 .
$$
Eliminating $t$,
$$
    v^2(t) = \sqrt{ v_0^2 + 2 a ( x(t)  - x_0 ) }
$$

You get

The position, $x(t)$, is

$$x(t) = x_0 + v_0 t + \frac{1}{2} a t^2 .$$

Eliminating $t$,

$$v^2(t) = \sqrt{ v_0^2 + 2 a ( x(t) - x_0 ) }$$

and

The angular momentum
\[
   \vec{L} = \vec{r} \times \vec{p}
\]

produces

The angular momentum

(1)$$\vec{L} = \vec{r} \times \vec{p}$$

The full equation syntax may be a little daunting, but you probably won’t need much more than the above. For the rest, there’s Markdown.

Oh, and you can draw pictures, using the built-in drawing tool.

(Just click on the “Create SVG Graphic”) button.

Forum: 362K – Topic: Notation for S.H.O.

Griffiths, Ch. 2 uses a slightly different notation for the raising and lowering operators in the 1D Harmonic Oscillator than I am. The dictionary is

Forum: 362K – Topic: Welcome

Here’s a place to discuss Physics 375R. You need to sign up (by clicking on the link at the upper left), before you can post.

Equations are pretty easy to type. If you type

The position, $x(t)$, is
$$
  x(t) = x_0 + v_0 t + \frac{1}{2} a t^2 .
$$
Eliminating $t$,
$$
    v^2(t) = \sqrt{ v_0^2 + 2 a ( x(t)  - x_0 ) }
$$

You get

The position, $x(t)$, is

$$x(t) = x_0 + v_0 t + \frac{1}{2} a t^2 .$$

Eliminating $t$,

$$v^2(t) = \sqrt{ v_0^2 + 2 a ( x(t) - x_0 ) }$$

and

The angular momentum
\[
   \vec{L} = \vec{r} \times \vec{p}
\]

produces

The angular momentum

(1)$$\vec{L} = \vec{r} \times \vec{p}$$

The full equation syntax may be a little daunting, but you probably won’t need much more than the above. For the rest, there’s Markdown.

Oh, and you can draw pictures, using the built-in drawing tool.

(Just click on the “Create SVG Graphic”) button.

Forum: 362L – Topic: HW 6 Typo?

Sigh. Yep, that’s a typo.

Forum: 362L – Topic: A Contour Integral

At some point, in Homework 3, you will encounter the integral

For those of you who took complex analysis, this should be easy to do, using the methods of contour integration.

If you haven’t taken complex analysis, the answer is

To see how the integral is done, read on…

First, let’s rewrite

The $i\epsilon$ displaces the location of the poles of the integrand off the real axis. The are located at $|k| +i\epsilon$ and at $-|k|-i\epsilon$. The contour can be closed either in the upper half-plane or in the lower half-plane. Closing in the upper half-plane, we pick up the pole at $|k| +i\epsilon$:

and we get $I= -\frac{i}{2|k|}$. If we closed the contour in the lower half-plane, we would pick up the pole at $-|k|-i\epsilon$. But, that contour is in the clockwise (instead of counter-clockwise) direction, the two minus-signs cancel, and we get the same result for the integral.

Forum: 362L – Topic: Lorentz Boost

I think you’re asking about how to relate Cartesian and spherical coordinates.

Our formula for Lorentz boosts was written in Cartesian coordinates. What we’re interested in is the relation between the corresponding systems of spherical coordinates.

Forum: 336K – Topic: Jordan Canonical Form

As we discussed, not every square matrix can be diagonalized, but every one can be put in Jordan canonical form

where $J$ is a block matrix of the form

where the $k^{\text{th}}$ Jordan block (of size $l_k\times l_k$) has the form

The special case, where all the Jordan blocks are $1\times 1$, is the case where $B$ is diagonalizable. When $B$ is diagonalizable, the matrix $S$ has $n$ columns, where the $k^{\text{th}}$ column is the eigenvector of $B$, corresponding to the eigenvalue ${\lambda }_k$:

or, equivalently,

Of course, this is a little bit ambiguous, as we can always multiply each $v_k$ by a non-zero constant. That ambiguity drops out of (1).

When a Jordan block has size $l_k>1$, we need, not one, but $l_k$ vectors to make up the corresponding columns of $S$. The first column is, again, given by the eigenvector

The next column is given by a vector, $w_{k,2}$, which satisfies

The column after that is given by $w_{k,3}$, which satisfies

and so on. Again, each of the $w_{k,i}$ is ambiguous by the addition of a multiple of $v_k$, but this drops out of (1).

Forum: 336K – Topic: Integral

In conversation at Office Hours, it came up that not every integral can be written in terms of elementary functions.

In particular, there’s a function called the “Error Function,”

Of course, we have that $\mathrm{Erf}(s)$ is a monotonic function, with $\mathrm{Erf}(0)=0$ and (more nontrivially) $\mathrm{Erf}(\mathrm{\infty })=1$.

To prove the latter, note that

Hope that helps …

Forum: 336K – Topic: Welcome

Here’s a place to discuss Physics 336K. You need to sign up (by clicking on the link at the upper left), before you can post.

Equations are pretty easy to type. If you type

The position, $x(t)$, is
$$
  x(t) = x_0 + v_0 t + \frac{1}{2} a t^2 .
$$
Eliminating $t$,
$$
    v^2(t) = \sqrt{ v_0^2 + 2 a ( x(t)  - x_0 ) }
$$

You get

The position, $x(t)$, is

$$x(t)=x_0+v_{0}t+\frac{1}{2}a{t}^2.$$x(t) = x_0 + v_0 t + \frac{1}{2} a t^2 .

Eliminating $t$,

$$v^2(t)=\sqrt{v_0^2+2a(x(t)-x_0)}$$v^2(t) = \sqrt{ v_0^2 + 2 a ( x(t) - x_0 ) }

and

The angular momentum
\[
   \vec{L} = \vec{r} \times \vec{p}
\]

produces

The angular momentum

(1)$$\stackrel{\rightharpoonup }{L}=\stackrel{\rightharpoonup }{r}\times \stackrel{\rightharpoonup }{p}$$\vec{L} = \vec{r} \times \vec{p}

The full equation syntax may be a little daunting, but you probably won’t need much more than the above. For the rest, there’s Markdown.

Oh, and you can draw pictures, using the built-in drawing tool.

(Just click on the “Create SVG Graphic”) button.

Forum: 302K – Topic: Ideal Gas Changes of State

Here is a table summarizing how various thermodynamic quantities change, when an ideal gas undergoes various processes.

The basic equations are

Here

is related to the molar specific heat at constant volume

and $\gamma =\frac{(\alpha +1)}{\alpha }$ is the ratio of specific heats

Forum: 302K – Topic: Static Equilibrium

Here, we’ll build the first stage of a house of cards. The two cards rest on a table with coefficient of static friction, ${\mu }_s$, as in the figure below.

The cards don’t fall, provided the total force and total torque on each card vanishes. By symmetry, we can examine the forces and torque on just one of the cards.

Let’s take the point of contact with the floor as the axis about-which to evaluate the torque. Then only two of the four forces (gravity and the force due to the other card, $N_c$) contribute to the torque. If we call the length of the card, $L$, the displacement at which gravity acts is $L/2$, and the perpendicular component of the gravitational force is $-mg\sin \theta$. The force from the other card acts at a displacement $L$, and the perpendicular component is $+N_c\cos \theta$. So the total torque is

The $x,y$ components of the total force are

We also must have

Combining (3) with (2), we have $N_c\le {\mu }_{s}mg$. Making the torque vanish in (1) requires

So we need

or

If we try an angle larger than that, the cards will slide.

Forum: 302K – Topic: Some Gravity Notes

In class, we considered only the case where one of the masses is much greater than the other, $M\gg m$, and so is more-or-less stationary. What happens when that isn’t the case?

Let’s consider the extreme opposite situtation, namely two equal masses, $m$, in a circular orbit about the midpoint between them.

The distance between them is $r=2R$, so the force on the first one is

(and, similarly, for $F_2$). So, for a circular orbit, we have

or

To compute the period, $T=\frac{2\pi R}{v}$, we get something that looks just like Kepler’s 3^rd Law.

but the constant is 4 times as big as before.

Forum: 302K – Topic: Some Gravity Notes

As we said in class, the acceleration due to gravity, near the Earth’s surface, is

The gravitational potential energy

where we’ve written $r=R_E+y$ and $y$ is the height above the Earth’s surface. For $\mid y\mid \ll R_E$, we can approximate this by a power series in $y$.

If you know Taylor series, you can compute it directly. If not, consider writing

for some choice of constants, $a_1,a_2,a_3,\dots$. If we multiply by $(R_E+y)$, we should get $1$:

So, to make the equation true, we must have

So (2) can be approximated

The first term is an irrelevant constant. The second is the gravitational potential energy we wrote down before, and the third term is the first correction, due to the fact that the gravitational force, due to the Earth, falls off with distance (rather than being constant).

Forum: 302K – Topic: Practice Exam(s)

I have gotten dozens of emails, asking questions about the Practice Exam. I’ll give the answers to those questions here.

1. No, the Practice Exam does not count for a grade.
2. Yes, it has a due-date (4 pm Sep 26, in the case of the 1^st one).
3. Yes, the solutions are available, as soon as the due-date has passed (just as they are for the homeworks).
4. No, I’m not going to change the due-date. The class seems to be divided between those who have taken the practice exam — and want to see the solutions — and those who haven’t taken it. The due-date was chosen so as to give everyone the maximum opportunity to do the practice exam, while still giving folks the opportunity to also study the solutions.

Update:

As it turns out, Quest had a bug (fixed now) which prevented students, who had not taken the practice exam before the due-date, from viewing the solutions. The bug is fixed now – alas, too late for the the 1^st exam.

Forum: 302K – Topic: Quest Troubles

Because the Quest website seems to be having major problems, I have delayed the due-date of 2^nd homework by 24 hours. I hope that they have things sorted out by then. Sorry for the inconvenience …

Forum: 302K – Topic: Disruption

I upgraded the software, on which this forum runs (to Rails 3.1.0 if you want to know the geeky details). The process didn’t go as smoothly as I would have liked, and service was pretty disrupted for most of Friday.

Should be back to normal now. But leave a comment here, if something’s still broken for you.

Forum: 302K – Topic: Discussion Sessions

Discussion sessions will start on Tuesday Sept 6.

Because Monday is Labour Day (and hence no discussion sessions that day), Jacob will hold an extra Tuesday session in ENS 116, from 6-7pm.

Forum: 302K – Topic: Dimensional Analysis

In class, we did two examples of dimensional analysis.

In the first, we deduced how $h(t)$, the height of water in the bucket, depended on $d$, the diameter of the bucket:

where $\tilde{k}$ is a constant (which depends on how fast the water is dripping from the tap) of dimensions $[L^3/T]$.

In the second example, we deduced that the period of a pendulum,

Can you think of other examples?

Forum: 302K – Topic: Welcome

Here’s a place to discuss Physics 302K. You need to sign up (by clicking on the link at the upper left), before you can post.

Equations are pretty easy to type. If you type

The position, $x(t)$, is
$$
  x(t) = x_0 + v_0 t + \frac{1}{2} a t^2 .
$$
Eliminating $t$,
$$
    v^2(t) = \sqrt{ v_0^2 + 2 a ( x(t)  - x_0 ) }
$$

You get

The position, $x(t)$, is

$$x(t)=x_0+v_{0}t+\frac{1}{2}a{t}^2.$$x(t) = x_0 + v_0 t + \frac{1}{2} a t^2 .

Eliminating $t$,

$$v^2(t)=\sqrt{v_0^2+2a(x(t)-x_0)}$$v^2(t) = \sqrt{ v_0^2 + 2 a ( x(t) - x_0 ) }

and

The angular momentum
\[
   \vec{L} = \vec{r} \times \vec{p}
\]

produces

The angular momentum

(1)$$\stackrel{\rightharpoonup }{L}=\stackrel{\rightharpoonup }{r}\times \stackrel{\rightharpoonup }{p}$$\vec{L} = \vec{r} \times \vec{p}

The full equation syntax may be a little daunting, but you probably won’t need much more than the above. For the rest, there’s Markdown.

Oh, and you can draw pictures, using the built-in drawing tool.

(Just click on the “Create SVG Graphic”) button.